A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly…

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability, that a student will get 4 or more correct answers just by guessing, is
  1. $\frac{10}{3^5}$
  2. $\frac{17}{3^5}$
  3. $\frac{13}{3^5}$
  4. $\frac{11}{3^5}$

Solution

Probability of guessing correct answer be $p=\frac{1}{3}$ $\therefore \quad \mathrm{q}=\frac{2}{3}$ Let random variable X denotes the number of correct answers. $\begin{array}{ll} \therefore \quad & X \sim B\left(5, \frac{1}{3}\right) \\ \therefore \quad & \text { Required probability }=\mathrm{P}(\mathrm{X} \geq 4) \\ & =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\ & ={ }^5 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)+{ }^5 \mathrm{C}_5\left(\frac{1}{3}\right)^5 \\ & =5 \times \frac{2}{3^5}+\frac{1}{3^5} \\ & =\frac{11}{3^5} \end{array}$

Asked in: MHT CET 2024 (04 May Shift 1)

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